竞赛优胜者的Mysql查询，包括在另一个表中有分数的抽奖条目

我正在我的网站上举办比赛。每场比赛都可以有多个参赛作品。我想根据分数在抽签的情况下检索最好的3个或更多条目。分数计算为所有投票的分数之和

以下是SQLFiddle：

表格条目如下所示：

id    contest_id
1     1          
2     1          
3     1          
4     1          
5     1          
6     1          
7     1          
8     1          
9     1

下表所列项目如下：

id    entry_id    score
-- entry 1 has 20 votes (5+10+5)
1     1           5
2     1           10
3     1           5
-- entry 2 has 20 votes (10+10)
4     2           10
5     2           10
-- entry 3 has 25 votes (5+5+5+10)
6     3           5
7     3           5
8     3           5
9     3           10
-- entry 4 has 10 votes (10)
10    4           10
-- entry 5 has 25 votes (10+10+5)
11    5           10
12    5           10
13    5           5
-- entry 6 has 5 votes (5)
14    6           5
-- entry 7 has 50 votes (10+10+10+10+10)
15    7           10
16    7           10
17    7           10
18    7           10
19    7           10
-- entry 8 has 20 votes (10+10)
20    8           10
21    8           10
-- entry 9 has 5 votes (5)
22    9           5

结果应与图纸一致：

id (entry_id) contest_id  score
7             1           50
3             1           25
5             1           25
1             1           20
2             1           20
8             1           20

我正在尝试以下查询：

select * from (
     select entry.*, sum(score) as final_score from entry join entry_vote on 
     entry.entry_id = entry_vote.entry_id group by entry_id
) as entries
where final_score in (
    select distinct(final_score) from (
        select entry_id, sum(score) as final_score from entry_vote group by entry_id
    ) as final_scores order by final_score desc limit 3;
)    

第一个子查询返回所有条目，所有条目的投票分数相加。 第二个子查询返回前三名的distincts总分50、25和20。

---

此查询返回错误。有什么问题以及如何解决？

您可以通过以下代码来解决

SELECT entry_id, SUM(score) AS score FROM entry_vote GROUP BY entry_id ORDER BY score DESC LIMIT 3;

---

以下是SQLFIDLE链接：

我设法通过使用联接而不是子查询来解决查询。 我在问题上发布的查询出现错误，因为第二个子查询的结尾是；：

select distinct(final_score) from (
    select entry_id, sum(score) as final_score from entry_vote group by entry_id
) as final_scores order by final_score desc limit 3;

解决这个问题，下一个错误将是子查询中的限制。Mysql不支持它们

最后一个问题是：

select * from (
    select entry.*, sum(score) as score from entry 
    join entry_vote on entry.entry_id = entry_vote.entry_id where entry.contest_id = <CONTEST_ID> group by entry_id
) as entry_with_score join (
    select distinct(score) as score from (
        select sum(score) as score from entry_vote 
        join entry on entry_vote.entry_id = entry.id 
        join contest on audition.contest_id = contest.id where contest.id = <CONTEST_ID> group by entry_id
    ) as score_by_entry order by score desc limit 3
) as top_score on entry_with_score.score = top_score.score; 

---

我想要前三名的条目。我已经编辑了SQL，它会给你前三名的分数。不，你不明白这个问题。这个查询是错误的，因为限制为3只会带来50分和25分，但正确的是分数为50分、25分和20分的条目。您的脚本不起作用，因为这个版本的MySQL还不支持第二个结果的“LIMIT&IN/ALL/ANY/SOME subquery”，你想得到哪一个1=20,2=20,8=20就告诉我我可以写一个存储过程。我忘了添加带有竞赛id的where子句。将更新答案！

 SELECT entry_id,total
   FROM 
      ( SELECT x.*
             , IF(@prev=total,@i,@i:=@i+1) i,@prev:=total prev 
          FROM 
             ( SELECT entry_id
                    , SUM(score) total
                 FROM entry_vote
                GROUP 
                   BY entry_id
             ) x
             , (SELECT @i:=0,@prev:=NULL)vars 
         ORDER 
            BY total DESC
      ) a
  WHERE i <=3;